LetXbe a vector space over the fieldKof real or complex numbers. We know from linear algebra that the (algebraic) dimension of X, denoted by dim(X), is the cardinality of a basis of X. If dim(X) is finite, we say that X isfinite dimensional otherwise X isinfinite dimensional. In this section we are going to focus on finite dimensional vector spaces.
Let{e1, . . . , ed}be a basis ofX, i.e. dim(X) =d. Given any vectorx∈X there exist unique x1, . . . , xd ∈ K s.t. x = x1e1 +· · ·+xded. This can be precisely expressed by saying that the mapping
Kd → X
(x1, . . . , xd) 7→ x1e1+· · ·+xded
is an algebraic isomorphism (i.e. linear and bijective) between X and Kd. In other words: IfX is a finite dimensional vector space then X is algebraically isomorphic to Kdim(X).
If now we give toX the t.v.s. structure and we considerK endowed with the euclidean topology, then it is natural to ask if such an algebraic isomor-phism is by any chance a topological one, i.e. if it preserves the t.v.s. structure.
The following theorem shows that ifX is a finite dimensional Hausdorff t.v.s.
then the answer is yes: X is topologically isomorphic toKdim(X).
It is worth to observe that usually in applications we deal always with Hausdorff t.v.s., therefore it makes sense to mainly focus on them.
Theorem 3.1.1. LetX be a finite dimensional Hausdorff t.v.s. overK(where K is endowed with the euclidean topology). Then:
a) X is topologically isomorphic to Kd, where d= dim(X).
b) Every linear functional on X is continuous.
c) Every linear map of X into any t.v.s. Y is continuous.
Before proving the theorem let us recall some lemmas about the continuity of linear functionals on t.v.s..
Lemma 3.1.2.
LetXbe a t.v.s. overKandv∈X. Then the following mapping is continuous.
ϕv : K → X ξ 7→ ξv.
Proof. For any ξ ∈ K, we have ϕv(ξ) = M(ψv(ξ)), where ψv : K → K×X given byψv(ξ) := (ξ, v) is clearly continuous by definition of product topology and M : K×X → X is the scalar multiplication in the t.v.s. X which is continuous by definition of t.v.s.. Hence, ϕv is continuous as composition of continuous mappings.
Lemma 3.1.3. Let X be a t.v.s. over K and L a linear functional on X.
Assume L(x)6= 0 for some x∈X. Then the following are equivalent:
a) L is continuous.
b) The null space Ker(L) is closed in X c) Ker(L) is not dense inX.
d) L is bounded in some neighbourhood of the origin in X, i.e. ∃ V ∈ F(o) s.t. supx∈V |L(x)|<∞.
Proof. (see Exercise Sheet 3) Proof. of Theorem 3.1.1
Let{e1, . . . , ed}be a basis of X and let us consider the mapping
ϕ: Kd → X
(x1, . . . , xd) 7→ x1e1+· · ·+xded.
As noted above, this is an algebraic isomorphism. Therefore, to conclude a) it remains to prove thatϕis also a homeomorphism.
Step 1: ϕis continuous.
When d= 1, we simply have ϕ ≡ ϕe1 and so we are done by Lemma 3.1.2.
When d > 1, for any (x1, . . . , xd) ∈ Kd we can write: ϕ(x1, . . . , xd) = A(ϕe1(x1), . . . , ϕed(xd)) = A((ϕe1 × · · · ×ϕed)(x1, . . . , xd)) where each ϕej
is defined as above andA:X×X→X is the vector addition in the t.v.s. X.
Hence, ϕis continuous as composition of continuous mappings.
Step 2: ϕis open and b) holds.
We prove this step by induction on the dimension dim(X) of X.
For dim(X) = 1, it is easy to see thatϕis open, i.e. that the inverse of ϕ:
ϕ−1 : X → K x=ξe1 7→ ξ is continuous. Indeed, we have that
Ker(ϕ−1) ={x∈X:ϕ−1(x) = 0}={ξe1 ∈X :ξ= 0}={o},
which is closed in X, since X is Hausdorff. Hence, by Lemma 3.1.3, ϕ−1 is continuous. This implies that b) holds. In fact, if L is a non-identically zero functional on X (when L ≡ 0, there is nothing to prove), then there exists a o 6= ˜x ∈ X s.t. L(˜x) 6= 0. W.l.o.g. we can assume L(˜x) = 1. Now for any x ∈ X, since dim(X) = 1, we have that x =ξx˜ for some ξ ∈K and so L(x) =ξL(˜x) =ξ. Hence,L≡ϕ−1 which we proved to be continuous.
Let d ∈ N with d > 1 and suppose now that both ϕ−1 is continuous and b) holds for dim(X) ≤ d−1. Let us first show that b) holds when dim(X) =d. LetL be a non-identically zero functional onX (when L ≡0, there is nothing to prove), then there exists a o 6= ˜x ∈ X s.t. L(˜x) 6= 0.
W.l.o.g. we can assume L(˜x) = 1. Note that for any x ∈ X the element x−xL(x)˜ ∈Ker(L). Therefore, if we take the canonical mapping φ :X → X/Ker(L) thenφ(x) =φ(˜xL(x)) =L(x)φ(˜x) for anyx∈X. This means that X/Ker(L) = span{φ(˜x)} i.e. dim(X/Ker(L)) = 1. Hence, dim(Ker(L)) = d−1 and so by inductive assumption Ker(L) is topologically isomorphic to Kd−1 1 This implies that Ker(L) is a complete subspace of X. Then, by Proposition 2.5.8-a),Ker(L) is closed in X and so by Lemma3.1.3we get L is continuous. By induction, we can conclude that b) holds whatever is the dimension ofX.
This immediately implies thatϕ−1 is continuous whatever isdim(X) (and so that a) holds for any dimension). In fact, the mapping
ϕ−1: X → Kd
x=Pd
j=1xjej 7→ (x1, . . . , xd) is continuous. Now for any x = Pd
j=1xjej ∈ X we can write ϕ−1(x) =
1Note that we can apply the inductive assumption not only because dim(Ker(L)) =d−1 but also becauseKer(L) is a Hausdorff t.v.s. since it is a linear subspace ofX which is an Hausdorff t.v.s..
(L1(x), . . . , Ld(x)), where for any j ∈ {1, . . . , d} we define Lj : X → K by Lj(x) := xjej. Since b) holds for any dimension, we know that each Lj is continuous and so ϕ−1 is continuous.
Step 3: The statement c) holds.
Let f : X → Y be linear and {e1, . . . , ed} be a basis of X. For any j ∈ {1, . . . , d} we define bj :=f(ej)∈ Y. Hence, for any x= Pd
j=1xjej ∈X we have f(x) =f(Pd
j=1xjej) =Pd
j=1xjbj. We can rewrite f as composition of continuous maps i.e. f(x) =A((ϕb1×. . .×ϕbd)(ϕ−1(x)) where:
• ϕ−1 is continuous by a)
• each ϕbj is continuous by Lemma 3.1.2
• Ais the vector addition on X and so it is continuous sinceX is a t.v.s..
Hence,f is continuous.
Corollary 3.1.4 (Tychonoff theorem). Let d ∈ N. The only topology that makes Kd a Hausdorff t.v.s. is the euclidean topology. Equivalently, on a finite dimensional vector space there is a unique topology that makes it into a Hausdorff t.v.s..
Proof.
We already know thatKd endowed with the euclidean topology τe is a Haus-dorff t.v.s. of dimension d. Let us consider another topology τ on Kd s.t.
(Kd, τ) is also Hausdorff t.v.s.. Then Theorem3.1.1-a) ensures that the iden-tity map between (Kd, τe) and (Kd, τ) is a topological isomorphism. Hence, as observed at the end of Section 1.1.4p.10, we get thatτ ≡τe.
Corollary 3.1.5. Every finite dimensional Hausdorff t.v.s. is complete.
Proof.
Let X be a Hausdorff t.v.s with dim(X) =d < ∞. Then, by Theorem 3.1.1-a),X is topologically isomorphic to Kdendowed with the euclidean topology.
Since the latter is a complete Hausdorff t.v.s., so isX.
Corollary 3.1.6. Every finite dimensional linear subspace of a Hausdorff t.v.s. is closed.
Proof.
Let S be a linear subspace of a Hausdorff t.v.s. (X, τ) and assume that dim(S) = d < ∞. Then S endowed with the subspace topology induced by τ is itself a Hausdorff t.v.s.. Hence, by Corollary 3.1.5 S is complete and therefore closed by Proposition 2.5.8-a).